Show that the function given by

The given function is $f(x)=\frac{\log x}{x}$.

$f^{\prime}(x)=\frac{x\left(\frac{1}{x}\right)-\log x}{x^{2}}=\frac{1-\log x}{x^{2}}$

Now, $f^{\prime}(x)=0$

$\Rightarrow 1-\log x=0$

$\Rightarrow \log x=1$

$\Rightarrow \log x=\log e$

$\Rightarrow x=e$

Now, $f^{\prime \prime}(x)=\frac{x^{2}\left(-\frac{1}{x}\right)-(1-\log x)(2 x)}{x^{4}}$

$=\frac{-x-2 x(1-\log x)}{x^{4}}$

$=\frac{-x-2 x(1-\log x)}{x^{4}}$

$=\frac{-3+2 \log x}{x^{3}}$

Now, $f^{\prime \prime}(e)=\frac{-3+2 \log e}{e^{3}}=\frac{-3+2}{e^{3}}=\frac{-1}{e^{3}}<0$

Therefore, by second derivative test, $f$ is the maximum at $x=e$.